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There are natural numbers: $a$, $b$, $c$.

$$\begin{cases} ab+bc+ca+\frac32(a+b+c)=5015,\\ 2abc-a-b-c=6366 \end{cases} $$

I need to find the minimum value of $a+b+c$. To my mind there's something connected with the derivative. I've already tried to find an equation for $a+b+c$, but I stuck right here. If we do some math in the second equation, we can get this one

$a+b+c=3183/abc$ Everything's great, but if we consider $a+b+c = l$ (for example), then we need to find a minimum of this function. But as far as I'm concerned this is a Hyperbolic function, which doesn't have extreme values. So there's my first confusion.

Moving on I decided to do something with a first equation. But, unfortunately, no matter how I rearrange my $a$, $b$ and $c$ I get some equations and can't get anything useful out of it.

And the last thing. If I get an equation for $a+b+c$ from the first equation then find a derivative and its zero value, I get $c = -b$.

Well, that perfectly concludes the problem. Could anyone help me get this one done?

Sorry for image confusion.

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The first equation in the system needs some check. Please do verify it. – user21436 Jan 18 '12 at 18:35
It's right. Checked it a dozen times – user1131662 Jan 18 '12 at 18:36
The second equation, when dividing both sides with 5, gives that $a+b+c=1003$ so $a+b+c$ is constant. Your problem is malformed, or very strange. – Per Alexandersson Jan 18 '12 at 18:41
I'm sorry, but maybe you're speaking abot the first one? – user1131662 Jan 18 '12 at 18:48
@user1131662 To be sure - the formulas I've edited into your post now are what you intended? – Martin Sleziak Jan 18 '12 at 19:05
up vote 3 down vote accepted

If you add the half the 2. equation to the 1. equation you get: $$ abc+(ab+bc+ca)+(a+b+c)=8198, $$ which can be rewritten as $$ (1+a)(1+b)(1+c) =8198+1=3\cdot 3 \cdot 911. $$ So $a=2,b=2$ and $c=910$.

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Wow, couldn't imagine it to be so simle as that. Nevertheless thank you for your spent time, it's really wonderful. And without derivatives. Again. – user1131662 Jan 18 '12 at 19:23
your welcome, I'm glad if could help – draks ... Jan 18 '12 at 19:34

Hint: Lagrange multipliers.

I mean, you've to find the minimum of the function

$$ f(a,b,c) = a + b + c \ , $$

subject to the restrictions $g_1(a,b,c) = 0$ and $g_2(a,b,c) = 0$, where

$$ g_1(a,b,c) = ab + bc + ca + \frac{3}{2}(a+b+c) - 5016 \qquad \text{and} \qquad g_2(a,b,c) = 2abc - a - b -c - 6366 \ . $$

So, you look for the zeros of the equation

$$ \nabla f(a,b,c) - \lambda_1 \nabla g_1(a,b,c) - \lambda_2 \nabla g_2(a,b,c) = 0 \ , $$

subject to the restrictions $g_1 = 0$ and $ g_2 = 0$.

Once you have those zeros $(a,b,c)$ together with its multipliers $\lambda_1, \lambda_2$, you have to check whether the Hessian at these points is a positive definite matrix or not:

$$ Hf(a,b,c) - \lambda_1 Hg_1 (a,b,c)- \lambda_2 Hg_2 (a,b,c) \ . $$

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Thanks, that needs a bit of studying, hopefully tomorrow I'll finish this thing. – user1131662 Jan 18 '12 at 19:53

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